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Appendix II.

P. 446. Culmann's Diagram.—To prove this construc-
tion refer to Fig. 818. Taking any section as B, we shall shew
that Bm = LU x H. Reaction x = G j, triangles K L M and o j G are

fV Q5     Q*    «*

U V   I   +=



similar, as also are N p M and o Q G ; and so on with c s P, D T s,
and E u T.

9.9 = p]^       and QGX/ = PMXH.. ...... = moment of 6.

H         p

Similarly    5 x/j = SP x H......... = moment of 5.

and    4 x/>2 = TS x H......... = moment of 4.

and    3X/3 = UTXH......... = moment of 3.

M L        _ ^ „,.,.„ _ M L x H = moment of x.

Again,   -    «

6       '


.'. Resultant moment = M4 - (M/6+ M/5 + M/4

= L u x H.

It is wise to call L u distance and H force, thus keeping
space and force diagrams quite separate. Shear is easily proved
from previous statements. (Seep. 1085.)

P. 4jo. Deflection of Beams by Graphic Summation.
—-Let a beam A v, Fig. 819, be bent to the curve A B c Y, by means
of a load of any kind, the beam section being of any shape, but
here considered uniform at all sections. Imagine a small portion
A B, in which the external fibres DD will be extended or compressed
by the amounts d and d± respectively; and if b = strain,